/*
Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = "great":

    great
   /    \
  gr    eat
 / \    /  \
g   r  e   at
           / \
          a   t
To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".

    rgeat
   /    \
  rg    eat
 / \    /  \
r   g  e   at
           / \
          a   t
We say that "rgeat" is a scrambled string of "great".

Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".

    rgtae
   /    \
  rg    tae
 / \    /  \
r   g  ta  e
       / \
      t   a
We say that "rgtae" is a scrambled string of "great".

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.
*/

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if (s1 == s2) return true;
        if (s1.size() != s2.size()) return false;
        vector<int> char_map(26, 0);
        int size = s1.size();
        for (int i = 0; i < size; i++) char_map[s1[i]-'a']++;
        for (int i = 0; i < size; i++) char_map[s2[i]-'a']--;
        for (int i = 0; i < char_map.size(); i++) { 
            if (char_map[i] != 0) return false; 
        } 
        for (int i=1; i<size;i++) {
            if ( (isScramble(s1.substr(0,i), s2.substr(0,i)) && isScramble(s1.substr(i), s2.substr(i))) ||
                 (isScramble(s1.substr(0,i), s2.substr(size-i)) && isScramble(s1.substr(i), s2.substr(0,size-i))) )
                return true;
        }
        return false;
    }
};
